Question

Integral only with u-sub method:

Answer 1

\[\int\limits_{0}^{1} \frac{ dx }{ (1+\sqrt{x})^4 }\] any hints?

Answer 2

\[u=1+\sqrt{x},\]\[du=(1/2\sqrt{x})dx\]\[2\sqrt{x}du=dx\]\[2udu=dx\]

Answer 3

and then do you limit changes, and then integrate

Answer 4

looks like I have to apply double integral subsbtitution

Answer 5

I did that yes :)

Answer 6

I'm working on it right now

Answer 7

x=1 u=2
x=0 u=1

Answer 8

\[\int\limits_{1}^{2}\frac{2u^2}{u^4}~du\]

Answer 9

then you can do it, I guess.

Answer 10

actually my u=sqrt(x)

Answer 11

no need

Answer 12

make it 1+sqrt{x}.
the derivative is same, and w/o any addition in denominator

Answer 13

u=1+sqrt{x} is much simpler, ain't it?

Answer 14

well, looks like it's going well so far lol

Answer 15

yes,
\[\int\limits_{1}^{2}\frac{2u^2}{u^4}~du=\int\limits_{1}^{2}2u^{-2}~du\]

Answer 16

from that point it is very simple, tell me the general anti derivative and then the antiderivative from 1 to 2.

Answer 17

what is the general antiderivative?

Answer 18

oh I made an error

Answer 19

\[\int\limits_{1}^{2}\frac{2u}{u^4}~du\]

Answer 20

because
du=1/2sqrt{x} dx
2sqrt{x} du= dx
2u du = dx

Answer 21

sorry for my mistake.
the u sub is good, but with one correction to my work

Answer 22

\[\int\limits_{1}^{2}\frac{2u}{u^4}~du\]

Answer 23

I get that for dx

Answer 24

yes, and I took the 2 out of the integral

Answer 25

yes, so my posted integral is your new expression

Answer 26

sure, whatever you like to do.

Answer 27

I'll post when I get an answer

Answer 28

\[2\int\limits_{1}^{2}u^{-3}~du\]

Answer 29

\[2\int\limits_{}^{}u^{-3}~du=2(-2)u^{-2}+C\]

Answer 30

go on, and beside the correct mistake, there are none mistakes

Answer 31

yep, my substitution got tricky :/